Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$ I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$
Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$
Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.
Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.
 A: Maple gives a rather unpleasant expression involving complex dilogarithm functions:
$$-\frac{19 \pi^{2}}{144}+\frac{7 \ln \! \left(2\right)^{2}}{12}-\frac{\ln \! \left(2-\sqrt{3}\right)^{2}}{6}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(-\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(\frac{1}{2}-\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}+\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(1-\frac{i \sqrt{3}}{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(1+\frac{i \sqrt{3}}{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}-\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}{3}
$$
I suspect this comes from writing the rational function $x^2/(1+x^3)$ in partial fractions and also writing $\ln(1-x^4) = \ln(1-x) + \ln(1+x) + \ln(1-ix) + \ln(1+ix)$.
